Mathematics

Solve $$\displaystyle\int\limits_0^{\dfrac{{\ln 3}}{2}} {\dfrac{{{e^x} + 1}}{{{e^{2x}} + 1}}dx} $$


ANSWER

$$\dfrac{\pi }{{12}} + \dfrac{1}{2}ln\left( {\dfrac{3}{2}} \right)$$


SOLUTION
$$\begin{array}{l}\displaystyle\int\limits_0^{\dfrac{{\ln 3}}{2}} {\dfrac{{{e^x} + 1}}{{{e^{2x}} + 1}}dx} \\\left[ \begin{array}{l}{e^x} = t\\{e^x}dx = dt\,\,\,\,\,\,\, \Rightarrow dx = \dfrac{{dt}}{t}\\x = 0\,\, \Rightarrow t = 1\\x = \dfrac{1}{2}\ln 3\,\, \Rightarrow t = {e^{\dfrac{1}{2}\ln 3}}\\t = \sqrt 3 \\
\end{array} \right]\\ = \displaystyle\int\limits_1^{\sqrt 3 } {\dfrac{{t + 1}}{{\left( {1 + {t^2}} \right)t}}} dt\\ = \displaystyle\int\limits_1^{\sqrt 3 } {\dfrac{{\left( {1 + \dfrac{1}{t}} \right)t}}{{\left( {1 + \dfrac{1}{{{t^2}}}} \right){t^2}}}\dfrac{{dt}}{t}} \\let\,\,\,\dfrac{1}{t} = u\\ - \dfrac{1}{{{t^2}}}dt = du\\ =  - \displaystyle\int\limits_1^{\dfrac{1}{{\sqrt 3 }}} {\dfrac{{1 + u}}{{1 + {u^2}}}du} \\ = \displaystyle\int\limits_{\dfrac{1}{{\sqrt 3 }}}^1 {\dfrac{1}{{1 + {u^2}}}du}  + \displaystyle\int\limits_{\dfrac{1}{{\sqrt 3 }}}^1 {\dfrac{u}{{1 + {u^2}}}du} \\ = \left( {{{\tan }^{ - 1}}\left( u \right)} \right)_{\dfrac{1}{{\sqrt 3 }}}^1 + \dfrac{1}{2}\left( {\ln \left( {{u^2} + 1} \right)} \right)_{\dfrac{1}{{\sqrt 3 }}}^1\\ = \dfrac{\pi }{{12}} + \dfrac{1}{2}\left( {\ln 2 - \ln \dfrac{4}{3}} \right)\\ = \dfrac{\pi }{{12}} + \dfrac{1}{2}ln\left( {\dfrac{3}{2}} \right)\end{array}$$
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Single Correct Medium Published on 17th 09, 2020
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